Question

If $\alpha$ and $\beta$ are the roots of $x^2+2x+2=0,$ then the value of ${\left(\frac{1}{\alpha +2}\right)}^3+{\left(\frac{1}{\beta +2}\right)}^3$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

$\alpha$ and $\beta$ are the roots of $x^2+2x+2=0$
$\Rightarrow {\alpha }^2+2\alpha +2=0$ and ${\beta }^2+2\beta +2=0$
$\Rightarrow \frac{1}{\alpha +2}=\frac{-\alpha }{2}$ and $\frac{1}{\beta +2}=\frac{-\beta }{2}$

Now, ${\left(\frac{1}{\alpha +2}\right)}^3+{\left(\frac{1}{\beta +2}\right)}^3$
$={\left(\frac{-\alpha }{2}\right)}^3+{\left(\frac{-\beta }{2}\right)}^3$
$=\frac{-1}{8}({\alpha }^3+{\beta }^3)$
$=\frac{-1}{8}(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)$
$=\frac{-1}{8}(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-3\alpha \beta ]$
$=\frac{-1}{8}\left(\frac{-2}{1}\right)\left[\right(-2{)}^2-3\times 2]$
$=\frac{-1}{2}$